After a golf ball is hit it takes off with an initial speed of 24.8 m/s and at a

Question

After a golf ball is hit it takes off with an initial speed of 24.8 m/s and at an angle of 41.5° with respect to the horizontal. The golf field is flat and horizontal.

a.) Neglecting air resistance how far will the golf ball fly?

b.)How high will the golf ball rise?

c.)How much time will the ball spend in the air? (3.35 s)<--(anwser)

d.)How far would the ball fly if the initial speed was doubled?

e.How much time would the ball spend in the air in this second case? (6.70 s)<--(anwser)

I just don't know the anwsers to the rest.

Explanation / Answer

Initial Horizontal speed = 24.8 * cos(41.5) = 18.57m/s
Initial Vertical speed = 24.8 * sin(41.5) = 16.43 m/s

Horizontal Acceleration = 0
Vertical Acceleration = 9.8 m/s^2 downwards

Max Height golf ball wiil rise =
V^2 = u^2 + 2as
At max height , Velocity =0
0 = 16.43^2 - 2*9.8*h
h = 13.77 m
Max Height golf ball wiil rise = h = 13.77 m

Total time spend by ball in air = Time taken to reach max height + Time taken to reach back ground
Time taken to reach max height =
t = u/a
t = 16.43/9.8 s
t = 1.676 s

This is the same time ball will take to reach back ground
Therefore ,
Total time spend by ball in air = 2 * 1.676s
Total time spend by ball in air = 3.35s

Distance travelled by Golf ball = Horizontal Velocity * Time spent in air
Distance travelled by Golf ball = 18.57 * 3.35 m
Distance travelled by Golf ball = 62.2 m

Now, If initial Speed is doubled -
New Horizontal speed = 2 * 24.8 * cos(41.5) = 37.14/s
New Vertical speed = 2 * 24.8 * sin(41.5) = 32.86 m/s

Total time spend by ball in air = 2 * 32.86/9.8 s = 2*3.35s
Total time spend by ball in air = 6.7 s

Distance travelled by Golf ball = Horizontal Velocity * Time spent in air
Distance travelled by Golf ball = 37.14 * 6.7 m
Distance travelled by Golf ball = 248.84 m

a.) 62.2 m
b.) 13.77 m
c.) 3.35 s
d.) 248.84 m
e.) 6.70 s

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